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Old July 3rd, 2009, 06:50 AM
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Default How to find a point?

Hello!

I'm trying to solve this problem:

Having k points in an n-dimentional space, I want to find a new point, any point P that is at distance 1 from each of them, or find out if it is impossible.
So, the system of equations would be

for each i = 1..k
Sum, for j = 1..n of (Xij - Pj) ^2 = 1

but Xij are constants, so

for each i = 1..k
Ci + (Sum, for j = 1..n of Pj ^2 - 2XijPj) = 1

With Ci = Sum, for j=1..n of Xij^2

Is there an easy, computational way of calculating this?

Sorry for my not-standard notation, I'm new to the forum.
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  #2  
Old July 3rd, 2009, 03:41 PM
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Quote:
Originally Posted by iluci View Post
Hello!

I'm trying to solve this problem:

Having k points in an n-dimentional space, I want to find a new point, any point P that is at distance 1 from each of them, or find out if it is impossible.
So, the system of equations would be

for each i = 1..k
Sum, for j = 1..n of (Xij - Pj) ^2 = 1

but Xij are constants, so

for each i = 1..k
Ci + (Sum, for j = 1..n of Pj ^2 - 2XijPj) = 1

With Ci = Sum, for j=1..n of Xij^2

Is there an easy, computational way of calculating this?

Sorry for my not-standard notation, I'm new to the forum.
Typically for a n-dimensional space, you will have a matrix with n dimensions.

suppose you have four points in a 5-dimensional space:

Point(x1,y1,z1,w1,v1)
Point(x2,y2,z2,w2,v2)
Point(x3,y3,z3,w3,v3)
Point(x4,y4,z4,w4,v4)

Then for your AnyPoint(x0,y0,z0,w0,v0)

for i= 1 to 5

Distance = \sqrt{(x_i-x_0)^2 + (y_i-y_0)^2 + (z_i-z_0)^2 + (w_i-w_0)^2 + (v_i-v_0)^2}

If Distance <> 1 then EXIT: NO SUCH NUMBER POSSIBLE.

next i

''AnyPoint() is at distance 1 from all other points.


If your number of points, exceeds the number of dimensions in your n-dimensional space, and NONE of the points are coincident with any other point then you will have an impossible condition to meet.

Just wondering, but how are you using more than three dimensions to define your points.



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Old July 3rd, 2009, 09:46 PM
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Originally Posted by aidan View Post

If your number of points, exceeds the number of dimensions in your n-dimensional space, and NONE of the points are coincident with any other point then you will have an impossible condition to meet.
If you are given three points in a plane, then it is possible(not is all cases) to find a point that is equidistant from all of them(the centre of the circle with three points on the circumference)..........it seems contradiction with your statement.
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Old July 4th, 2009, 06:36 AM
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If you are given three points in a plane, then it is possible(not is all cases) to find a point that is equidistant from all of them(the centre of the circle with three points on the circumference)..........it seems contradiction with your statement.
malaygoel, you are correct.

I'm sorry, but I phrased that badly.
When my teacher slapped some sense into me, my train of thought was derailed.

A sphere (centered in a cube) will intersect each face of the cube resulting in a circle of points. All of the points on the 6 circles will be at the sphere's radius distance.

Instead of sinking into gibberish, I'll find the correct phrasing and post it.

Thanks for the wake up.
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Old July 4th, 2009, 07:10 AM
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aidan, thanks for the help, but what you were suggesting was _checking_ if a point fits the description. I need to calculate the coordonates of such a point since I can't just pick points at random and hope I get it right

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Just wondering, but how are you using more than three dimensions to define your points.
I don't think I understood the question.
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Old July 4th, 2009, 08:45 AM
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...I don't think I understood the question.
I'm not sure either.

Can you give two(2) sets of n-dimensional data?

That way I will understand exactly what you are posing.

Give 1 set (at least half-dozen points) in 4-dimensional space and
create a 2nd set (again with about six points) in 7 dimensional space.

It appears that you are looking for an unknown point equidistant from all other points in the specific set of points.

Please clarify.
Give me the concrete data, and I'll should you the routine for a resolution.
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